
John M. answered 03/22/14
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Analytical assistance -- Writing, Math, and more
Sun
For part (a), remember that velocity is the derivative of position with respect to time, and acceleration is the derivative of velocity with respect to time. These derivatives can be performed separately and combined through vector addition. So we are given dx/dt=4t+1, and dy/dt=sin(t^2). At t=3, dx/dt =13 and dy/dt=sin(9)=.412, and using the Pythagorean theorem produces 13.0065. You can repeat this process for acceleration by taking the derivative of both dx/dt and dy/dt with respect to t, compute the accelerations in each direction and then compute their combined vector acceleration.
For part (b), the slope of the tangent line is dy/dx = (dy/dt)/(dx/dt), and then substitute 3 for t.
For part (c), if you integrate dx/dt, you get x=2t^2+t+C, solving for C using x(0)=0, gives C=0. The problem is that you cannot repeat this for dy/dt because sin(t^2) cannot be easily integrated, but it can be estimated by using taylor series (which I assume you have access to). If you integrate substituting the taylor series for sine, you can solve for its C (using y(0)=-4, and then solve both equations to find (x,y) at t=3.
For part (d), you need to calculate the arc length. The arc length (L) equals ∫ds, where
ds=√((dy/dt)^2+(dx/dt)^2) dt
ds=√((dy/dt)^2+(dx/dt)^2) dt
from t=0 to t=3.
A good description of parametric equations and how to use them can be found at
And the formula for arc lengths using parametric equations
I hope this points you in the right direction.
Sun K.
03/23/14